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if \: lx + my + n = 0 \: is \: the \: tangent \: of \: {x }^(2) + {y }^(2) = {a}^(2)\: then \: what \: will \: be \: the \: point \: of \: contact {?}



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Answer:

(x, y) = (-ln/(l²+m²), -mn/(l²+m²))

Explanation:

The given equation for the tangent line can be written in standard form as ...

lx +my = -n

The perpendicular line through the origin (the center of the circle) is ...

mx -ly = 0

The point of intersection of these lines is the point of tangency. It can be found a number of ways. Cramer's rule offers perhaps the simplest:

(x, y) = (l, m)×(-n/(l²+m²))

(x, y) = (-ln/(l²+m²), -mn/(l²+m²))

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In order for this point of tangency to be on the circle, we must have ...

a² = n²/(l²+m²)

which means another way to write the coordinates of the tangent point is ...

(x, y) = (-a²l/n, -a²m/n)

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Attached is an example of the problem and the solution.

if \: lx + my + n = 0 \: is \: the \: tangent \: of \: {x }^(2) + {y }^(2) = {a}^(2)\: then-example-1
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